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If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is
If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?
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Practice Questions
1 question
Q1
If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?
3V
6V
9V
4.5V
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Using the voltage divider rule, V6Ω = (R6Ω / (R3Ω + R6Ω)) * Vtotal = (6 / (3 + 6)) * 9 = 6V.
Questions & Step-by-step Solutions
1 item
Q
Q: If a circuit has a 9V battery and two resistors of 3Ω and 6Ω in series, what is the voltage across the 6Ω resistor?
Solution:
Using the voltage divider rule, V6Ω = (R6Ω / (R3Ω + R6Ω)) * Vtotal = (6 / (3 + 6)) * 9 = 6V.
Steps: 7
Show Steps
Step 1: Identify the total voltage from the battery, which is 9V.
Step 2: Identify the resistors in the circuit. We have a 3Ω resistor and a 6Ω resistor in series.
Step 3: Calculate the total resistance in the circuit by adding the two resistors together: 3Ω + 6Ω = 9Ω.
Step 4: Use the voltage divider rule to find the voltage across the 6Ω resistor. The formula is: V6Ω = (R6Ω / (R3Ω + R6Ω)) * Vtotal.
Step 5: Substitute the values into the formula: V6Ω = (6 / (3 + 6)) * 9.
Step 6: Simplify the equation: V6Ω = (6 / 9) * 9.
Step 7: Calculate the result: V6Ω = 6V.
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